Boundaries of finite intersections and unions of sets I apologize if this is a duplicate - I looked but didn't find one.
This question is sort of a sanity check.
Let $A$, $B$ be sets and define the boundaries $\partial A$ and $\partial B$ as usual.
Is it true that both  $\partial (A \cup B) \subseteq \partial A \cup \partial B$ and $\partial (A \cap B) \subseteq \partial A \cup \partial B$?
It seems obvious, and the proofs seem really easy, but I haven't seen this fact written down anywhere.
To get started on the proofs I thought to look at a point $p$ which is in neither $\partial A$ nor  $\partial B$ and then show it is not in either $\partial (A \cup B)$ or $\partial (A \cap B)$ by looking at different cases where $p$ is in the interior or exterior of $A$, $B$, taking intersections of neighborhoods, etc...
Thanks a bunch! 
 A: $\newcommand{\bdry}{\operatorname{bdry}}\newcommand{\cl}{\operatorname{cl}}\newcommand{\int}{\operatorname{int}}$You can do it with inline calculations if you use the definition that $\bdry A=\cl A\cap\cl(X\setminus A)$:
$$\begin{align*}
\bdry(A\cap B)&=\cl(A\cap B)\cap\cl\Big(X\setminus(A\cap B)\Big)\\
&=\cl(A\cap B)\cap\cl\Big((X\setminus A)\cup(X\setminus B)\Big)\\
&=\cl(A\cap B)\cap\Big(\cl(X\setminus A)\cup\cl(X\setminus B)\Big)\\
&=\Big(\cl(A\cap B)\cap\cl(X\setminus A)\Big)\cup\Big(\cl(A\cap B)\cap\cl(X\setminus B)\Big)\\\\
&\subseteq\Big(\cl A\cap\cl(X\setminus A)\Big)\cup\Big(\cl B\cap\cl(X\setminus B)\Big)\\\\
&=\bdry A\cup\bdry B\;,
\end{align*}$$
and
$$\begin{align*}
\bdry(A\cup B)&=\cl(A\cup B)\cap\cl\Big(X\setminus(A\cup B)\Big)\\
&=\cl(A\cup B)\cap\cl\Big((X\setminus A)\cap(X\setminus B)\Big)\\
&\subseteq\cl(A\cup B)\cap\Big(\cl(X\setminus A)\cap\cl(X\setminus B)\Big)\\
&=\Big(\cl A\cup\cl B\Big)\cap\cl(X\setminus A)\cap\cl(X\setminus B)\\
&=\left(\Big(\cl A\cap\cl(X\setminus A)\Big)\cup\Big(\cl B\cap\cl(X\setminus A)\Big)\right)\cap\cl(X\setminus B)\\
&=\Big(\bdry A\cap\cl(X\setminus B)\Big)\cup\Big(\cl B\cap\cl(X\setminus A)\cap\cl(X\setminus B)\Big)\\
&=\Big(\bdry A\cap\cl(X\setminus B)\Big)\cup\Big(\bdry B\cap\cl(X\setminus A)\Big)\\\\
&\subseteq\bdry A\cup\bdry B\;.
\end{align*}$$
